34 2. Analytic Functions - Mathyplee/teaching/4200f18/Ch2_pages.pdfexponential function, for...

34 2. Analytic Functions

6. Find the interior, closure, and boundary for the set {z ∈ C : 1 ≤ |z| < 2} (noproof required).

7. Prove that w ∈ C is in the closure of a set E ⊂ C if and only if there is asequence {zn} ⊂ E such that lim zn = w. Thus, a set E is closed if and only ifit contains all limits of convergent sequences of points in E.

8. Does limz→0 f(z) exist if f(z) =|z − z|

|z| with domain C \ {0}? How about if

the domain is restricted to be just R \ {0}?

9. Prove that Re(z), Im(z), and z are continuous functions of z.

10. At which points of C is the function (1 − z4)−1 continuous.

11. Prove that argI is continuous except on its cut line.

12. Use the result of the preceding exercise to prove that a branch of the log functionis continuous except on its cut line.

13. Use Theorem 2.1.13 to prove that if f and g are continuous functions with opendomains Uf and Ug and if g(Ug) ⊂ Uf , then f ◦ g is continuous on Ug.

14. Prove that if f is a continuous function defined on an open subset U of C, thensets of the form {z ∈ U : |f(z)| < r} and {z ∈ U : Re(f(z)) < r} are open.

15. Use the result of the preceding exercise to come up with an open subset of Cthat has not been previously described in this text.

16. Prove that a function f with open domain U is continuous at a point a ∈ Uif and only if whenever {zn} ⊂ U is a sequence converging to a, the sequence{f(zn)} converges to f(a).

2.2. The Complex Derivative

There is nothing surprising about the definition of the derivative of a function ofa complex variable – it looks just like the definition of the derivative of a functionof a real variable. What is surprising are the consequences of a function having aderivative in this sense.

Definition 2.2.1. Let f be a function defined on a neighborhood of z ∈ C. If

limw→z

f(w) − f(z)

w − z

exists, then we denote it by f ′(z) and we say f is differentiable at z with complexderivative f ′(z). If f is defined and differentiable at every point of an open set U ,then we say that f is analytic on U .

Remark 2.2.2. When convenient, we will make the change of variables λ = w− zand write the derivative in the form

(2.2.1) f ′(z) = limλ→0

f(z + λ) − f(z)

λ.

Clearly constant functions are differentiable and have complex derivative 0,since the difference quotient in Definition 2.2.1 is identically 0 for such a function.

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2.2. The Complex Derivative 35

The first hint that there is something fundamentally different about this notionof derivative is in the following example.

Example 2.2.3. Show that the function f(z) = z is differentiable everywhere onC with derivative 1 and, hence, is analytic on C, but the function f(z) = z isdifferentiable nowhere.

Solution: For f(z) = z, the difference quotient in (2.2.1) is

λ

λ= 1,

which clearly has limit 1 as λ → 0 for every z. On the other hand, if f(z) = z, thenthe difference quotient is

λ

λ= e−2iθ,

if λ = r eiθ in polar form. The limit of this function as λ → 0 clearly does not exist,since it has a different fixed value along each ray emanating from 0. This is trueno matter what z is, and so z is nowhere differentiable.

What makes this example so surprising, at first, is that, as a function of thetwo real variables x and y, z = x − iy is of class C∞ – meaning that its partialderivatives of all orders exist and are continuous – and yet, its complex derivativedoes not exist. Thus, existence of the complex derivative involves more than justsmoothness of the function.

We will soon prove that a function which has a power series expansion thatconverges on an open disc is analytic on that disc. This would imply that theexponential function, for example, is analytic on all of C. We do not have to wait,however, to prove this fact. There is an elementary proof that ez is analytic on C.

Example 2.2.4. Prove that ez is an analytic function of z on the entire complexplane and show that it is its own derivative.

Solution: Given an arbitrary point z ∈ C, we will show that ez has derivativeez at z. By the law of exponents

ez+λ − ez

λ= ez eλ −1

λ.

Thus, to show that the derivative of ez is ez we need only show that

(2.2.2) limλ→0

eλ −1

λ= 1.

However, if t = |λ|, inspection of the power series for eλ and et shows that

(2.2.3)

∣∣∣∣eλ −1

λ− 1

∣∣∣∣ =

∣∣∣∣eλ −1 − λ

λ

∣∣∣∣ ≤et −1 − t

t.

Now to show that the expression on the left has limit zero and, thus, verify (2.2.2),we simply apply L’Hopital’s rule to the expression on the right.



Elementary Properties of the Derivative. A simple result about derivativesof functions of a real variable that also holds in the context of complex derivativesis the following. The proof is elementary and is left to the exercises.

Theorem 2.2.5. If the complex derivative f ′ of f exists at a ∈ C, then f iscontinuous at a.

The complex derivative has all of the familiar properties in relation to sums,products, and quotients of functions. The proofs of these are in no way differentfrom the proofs of the corresponding results for functions of a real variable. In thefollowing theorem, Part (a) is trivial and we leave Parts (b) and (c) to the exercises.

Theorem 2.2.6. If f and g are functions of a complex variable which are differ-entiable at z ∈ C, then

(a) f + g is differentiable at z and (f + g)′(z) = f ′(z) + g′(z);

(b) fg is differentiable at z and (fg)′(z) = f ′(z)g(z) + f(z)g′(z);

(c) if g(z) = 0, 1/g is differentiable at z and (1/g)′(z) = −g′(z)/g2(z).

Parts (a) and (b) of this theorem and the fact that constant functions and thefunction z are analytic on C imply that every polynomial in z is analytic on C. Ofcourse, since z is not analytic, we cannot expect mixed polynomials that containpowers of both z and z to be analytic.

Parts (b) and (c) of the theorem imply that f/g is differentiable at z if f andg are and if g(z) = 0. They also imply the quotient rule

(f

g

)′(z) =

f ′(z)g(z) − g′(z)f(z)

g2(z).

The chain rule also holds for the complex derivative.

Theorem 2.2.7. If g is differentiable at a and f is differentiable at b = g(a), thenf ◦ g is differentiable at a and

(f ◦ g)′(a) = f ′(g(a))g′(a).

Proof. Let U be a neighborhood of b on which f is defined. We define a functionh(w) on U in the following way

h(w) =

⎧⎨

⎩

f(w) − f(b)

w − b, if w = b;

f ′(b), if w = b.

Then h is continuous at b, since

f ′(b) = limw→b

f(w) − f(b)

w − b.

Also,

(2.2.4)f ◦ g(z) − f ◦ g(a)

z − a= h(g(z))

g(z) − g(a)

z − a

for all z in the deleted neighborhood V = g−1(U) \ {a} of a. If we take the limitof both sides of (2.2.4) and use the fact that f and h are continuous at b and g iscontinuous at a, we conclude that (f ◦ g)′(a) = f ′(g(a))g′(a), as required. !



Example 2.2.8. Suppose p(z) is a polynomial in z. Where is the function ep(z)

analytic and what is its derivative?

Solution: Since ez and p(z) both are differentiable everywhere, so is the com-position ep(z), by Theorem 2.2.7, and the derivative is

(ep(z)

)′= p′(z) ep(z) .

The Cauchy-Riemann Equations. Since a function f of a complex variablemay be regarded as a complex-valued function on a subset of R2, we can write itin the form

(2.2.5) f(x + iy) = u(x, y) + iv(x, y),

where u and v are the real and imaginary parts of f , regarded as functions definedon a subset of R2. It is natural to ask what the existence of a complex derivativefor f implies about the functions u and v as functions of the two real variables xand y. It is easy to see that it implies the existence of the partial derivatives ux,uy, vx and vy. In fact, it implies much more as the following discussion will show.

Recall that a function g of two real variables is said to be differentiable at (x, y)if there are numbers A and B such that

g(x + h, y + k) − g(x, y) = Ah + Bk + ϵ(h, k),

where ϵ(h, k)/|(h, k)| → 0 as (h, k) → (0, 0). If g is differentiable at (x, y), then thenumbers A and B are the partial derivatives gx and gy at (x, y).

Suppose f is a complex-valued function defined in a neighborhood of z ∈ C. IfM = f ′(z) exists, then we may write

(2.2.6) f(z + λ) − f(z) = Mλ + ϵ(λ),

where ϵ(λ)/λ → 0 as λ → 0. In fact, ϵ(λ) is given by

ϵ(λ) = f(z + λ) − f(z) − Mλ,

and so, the fact that ϵ(λ)/λ → 0 as λ → 0 is equivalent to the statement that f ′(z)exists and is equal to M .

If we write f, M, z,λ, and ϵ in terms of their real and imaginary parts: f =u + iv, M = C + iD, z = x + iy,λ = h + ik, and ϵ = ρ + iω, then (2.2.6) becomes

(2.2.7) u(x + h, y + k) + iv(x + h, y + k) − u(x, y) − iv(x, y)

= (C + iD)(h + ik) + ρ(h, k) + iω(h, k).

On equating real and imaginary parts, this leads to the two equations

u(x + h, y + k) − u(x, y) = Ch − Dk + ρ(h, k),

v(x + h, y + k) − v(x, y) = Dh + Ck + ω(h, k).(2.2.8)

The condition that ϵ(λ)/λ → 0 as λ → 0 implies that ρ(h, k)/|(h, k)| → 0 andω(h, k)/|(h, k)| → 0 (note that |(h, k)| =

√h2 + k2 = |λ|). Thus, we can draw two

conclusions from the existence of f ′(z): (1) u and v are differentiable at (x, y), and(2) the partial derivatives of u and v at (x, y) are given by

ux(x, y) = C, uy(x, y) = −D,

vx(x, y) = D, vy(x, y) = C.(2.2.9)



A surprising consequence of this is that if f ′ exists at z = x + iy, then

ux = vy,

uy = −vx(2.2.10)

at (x, y). Equations (2.2.10) are the Cauchy-Riemann equations . Equations (2.2.9)also show that if f ′ exists at z, then f ′(z) = C + iD = ux + ivx = −i(uy + ivy). Ifwe set fx = ux + ivx and fy = uy + ivy, then this can be written as f ′ = fx = −ify

wherever f ′ exists.

The above discussion shows that, at any point where f has a complex derivative,its real and imaginary parts are differentiable functions and satisfy the Cauchy-Riemann equations. The converse is also true: If the real and imaginary parts of fare differentiable and satisfy the Cauchy-Riemann equations at a point z = x + iy,then f ′(z) exists. The proof of this is a matter of working backwards through theabove discussion, beginning with the assumption that u and v are differentiable at(x, y), with partial derivatives that satisfy ux = vy = C and uy = −vx = −D. Thisleads to (2.2.8), which eventually leads back to the conclusion that C + iD is thederivative of f at z = x + iy. We leave the details to the exercises. The result isthe following theorem.

Theorem 2.2.9. If f = u + iv is a complex-valued function defined in a neigh-borhood of z ∈ C, with real and imaginary parts u and v, then f has a complexderivative at z if and only if u and v are differentiable and satisfy the Cauchy-Riemann equations (2.2.10) at z = x + iy. In this case,

f ′ = fx = −ify.

Example 2.2.10. We already know that ez is analytic everywhere. However, givea different proof of this by showing ez satisfies the Cauchy-Riemann equations.

Solution: With z = x + iy, we write ez = ex(cos y + i sin y). The real andimaginary parts of ez are u(x, y) = ex cos y and v(x, y) = ex sin y. Thus,

ux(x, y) = ex cos y = vy, and

uy(x, y) = − ex sin y = −vx.

Example 2.2.11. Use the Cauchy-Riemann equations to prove that, for eachbranch of the log function, log(z) is analytic everywhere except on its cut lineand has derivative 1/z.

Solution: We first prove that the principal branch of the log function is an-alytic on the right half-plane H = {z ∈ C : Re(z) > 0}. For z ∈ H we havez = x + iy = r eiθ where

r =√

x2 + y2 and θ = tan−1(y/x).

Thus, the principal branch of log on H is

log(x + iy) = (1/2) ln(x2 + y2) + i tan−1(y/x).



Taking partial derivatives yields

∂

∂x(1/2) ln(x2 + y2) =

x

x2 + y2,

∂

∂xtan−1(y/x) =

−y/x2

1 + (y/x)2=

−y

x2 + y2,

∂

∂y(1/2) ln(x2 + y2) =

y

x2 + y2,

∂

∂ytan−1(y/x) =

1/x

1 + (y/x)2=

x

x2 + y2.

(2.2.11)

Thus, the Cauchy-Riemann equations are satisfied by the principal branch of thelog function on H. Furthermore

(log z)′ =∂

∂xlog(x + iy) =

x − iy

x2 + y2=

1

z.

Now if z is any point not on the negative real axis and not in H, then we simplyrotate z into H. That is, we choose α = ±π/2 such that eiα z ∈ H. Then

log z = log(eiα z) − iα.

Since log has derivative 1/w at w = eiα z, it follows from the chain rule that log hasderivative eiα /(eiα z) = 1/z at z. Thus, the principal branch of the log function isanalytic with derivative 1/z at any point z not on its cut line.

The analogous statement for other branches of the log function also followsfrom a rotation argument, as above. That is, each such function is just the principalbranch of the log function composed with a rotation.

Harmonic Functions. In the next chapter, we will prove that analytic functionsare C∞– that is, they have continuous complex derivatives of all orders. This, inparticular, implies that analytic functions have continuous partial derivatives of allorders with respect to x and y. Assuming this result for the moment, we have

Theorem 2.2.12. The real and imaginary parts of an analytic function on U areharmonic functions on U , meaning they satisfy Laplace’s equation

uxx + uyy = 0.

Proof. If f = u + iv is an analytic function, then u and v satisfy the Cauchy-Riemann equations and so

uxx = (ux)x = (vy)x = (vx)y = (−uy)y = −uyy.

This shows that the real part of f satisfies Laplace’s equation. Since v is the realpart of the analytic function −if , it follows that v is also harmonic. Thus, bothreal and imaginary parts of an analytic function are harmonic. !

If u and v are harmonic functions such that the function f = u+ iv is analytic,then we say u and v are harmonic conjugates of one another.

Example 2.2.13. Prove that u(x, y) = ex cos y is a harmonic function on all of R2

and find a harmonic conjugate for it.



Solution: The function u is the real part of f(z) = ez and is, therefore,harmonic by the previous theorem. The imaginary part of f is v(x, y) = ex sin y,and so this function v is a harmonic conjugate of u.

Exercise Set 2.2

1. Fill in the details in Example 2.2.4 by verifying the inequality (2.2.3) andshowing that the limit of the expression on the right is 0.

2. Prove Theorem 2.2.5.

3. Prove Part (b) of Theorem 2.2.6.

4. Prove Part (c) of Theorem 2.2.6.

5. Use induction and Theorem 2.2.6 to show that (zn)′ = nzn−1 if n is a non-negative integer.

6. Find the derivative of z7 + 5z4 − 2z3 + z2 − 1. Which results from this sectionare used in this calculation?

7. Find the derivative of ez3.

8. If we use the principal branch of the log function, at which points of C doeslog z

zhave a complex derivative? What is its derivative at these points?

9. Finish the proof of Theorem 2.2.9 by showing that if f = u + iv, u and v aredifferentiable at z, and u and v satisfy the Cauchy-Riemann equations at z,then f ′(z) exists.

10. Use the Cauchy-Riemann equations to verify that the function f(z) = z2 isanalytic everywhere.

11. Describe all real-valued functions which are analytic on C.

12. Derive the Cauchy-Riemann equations in polar coordinates:

ur = r−1vθ,

uθ = −rvr

by using the change of variable formulas x = r cos θ, y = r sin θ and the chainrule.

13. We showed in Example 2.2.11 that each branch of the log function is analyticon the complex plane with its cut line removed. Use the Cauchy-Riemannequations in polar form (previous problem) to give another proof of this fact.

14. Assuming each branch of the log function is analytic, use the chain rule to giveanother prove that each such function has derivative 1/z.

15. Use the Cauchy-Riemann equations to prove that if f is analytic on an open setU , then the function g defined by g(z) = f(z) is analytic on the set {z : z ∈ U}.

16. Verify that the function log |z| is harmonic on C\{0} and find a harmonic con-jugate for it on the set consisting of C with the non-positive real axis removed.


2.3. Contour Integrals 41

2.3. Contour Integrals

Integration plays a key role in this subject – specifically, integration along curvesin C. A curve or contour in the plane C is a continuous function γ from an intervalon the line into C. Such an object is sometimes called a parameterized curve andthe interval I is called the parameter interval. We will be interested in a particularkind of curve, one whose parameter interval is a closed bounded interval which canbe subdivided into finitely many subintervals, on each of which γ is continuouslydifferentiable.

Smooth Curves. Let I = [a, b] be a closed interval on the real line and letγ : I → C be a complex-valued function on I. If c ∈ I, then the derivative γ′(c) ofγ at c is defined in the usual way:

(2.3.1) γ′(c) = limt→c

γ(t) − γ(c)

t − c.

Of course, γ is complex-valued and so this limit should be interpreted as the typeof limit discussed in Section 2.1. It can be calculated by expressing γ in terms ofits real and imaginary parts, that is, by writing γ(t) = x(t) + iy(t), where x(t) andy(t) are real-valued functions on I. Then γ′(t) = x′(t) + iy′(t) (Exercise 2.3.6).

What about the endpoints a and b of the interval I? Should we either not talkabout the derivative at the endpoints or, perhaps, use one-sided derivatives definedin terms of one-sided limits (limit from the right at a and limit from the left at b)?Actually, there is no need to do anything special at a and b or to exclude them. Ifthe domain of γ is [a, b], then our domain dependent definition of limit takes careof the problem. If c = a, the limit as t → a in (2.3.1) only involves values of t tothe right of a, since only those are in the domain of the difference quotient thatappears in this limit. Similarly, if c = b, the limit as t → b involves only points tothe left of b. Thus, the derivatives at a and b that our definition leads to are whatin calculus would be called the right derivative at a and the left derivative at b.

The curve γ is differentiable at c if the limit defining γ′(c) exists. It is contin-uously differentiable or smooth on I if it is differentiable at every point of I and ifthe derivative is a continuous function on I. In this case we will write γ ∈ C1(I).

Definition 2.3.1. A curve γ : [a, b] → C in C is called piecewise smooth if thereis a partition a = a0 < a1 < · · · < an = b of [a, b] such that the restriction of γ to[aj−1, aj ] is smooth for each j = 1, · · · , n. A curve which is piecewise smooth willbe called a path.

With appropriate choices of parameterization, familiar geometric objects in Ccan be described as the image of a path.

Example 2.3.2. Find a path γ that traces once around the circle of radius r,centered at 0, in the counterclockwise direction. Describe γ′.

Solution: The smooth path γ(t) = r eit, t ∈ [0, 2π] does the job. Its derivativemay be obtained by writing it as r(cos t + i sin t) and differentiating the real andimaginary parts. The result is γ′(t) = r(− sin t + i cos t) = ir eit.



γ(a)

γ(b)

γ(t)

.

..

Figure 2.3.1. A Path in the Plane.

Example 2.3.3. Let z and w be two points in C. Find a path which traces thestraight line from z to w and find its derivative.

Solution: The path γ, with parameter interval [0, 1], defined by

γ(t) = (1 − t)z + tw = z + t(w − z),

satisfies γ(0) = z and γ(1) = w. It is a parametric form of a straight line in theplane, and its derivative is γ′(t) = w − z.

Example 2.3.4. Find a path that traces once around the square with vertices0, 1, 1 + i, i in the counterclockwise direction. Find γ′(t) on the subintervals whereγ is smooth.

Solution: We choose [0, 1] as the parameter interval and define a path γ asfollows (see Figure 2.3.2):

γ(t) =

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

4t, if 0 ≤ t ≤ 1/4;

1 + (4t − 1)i, if 1/4 ≤ t ≤ 1/2;

3 − 4t + i, if 1/2 ≤ t ≤ 3/4;

(4 − 4t)i, if 3/4 ≤ t ≤ 1.

This is continuous on [0, 1] and smooth on each subinterval in the partition 0 <1/4 < 1/2 < 3/4 < 1. It traces each side of the square in succession, moving inthe counterclockwise direction. On the first interval, γ′ is the constant 4, on thesecond it is 4i, on the third it is −4, and on the fourth it is −4i.

Riemann Integral of Complex-Valued Functions. The integral of a functionalong a path will be defined in terms of the Riemann integral on an interval. Thisis the familiar Riemann integral from calculus, except that the functions beingintegrated will be complex-valued. This difference requires a few comments.

If f(t) = g(t) + ih(t) is a complex-valued function on an interval [a, b], whereg and h are real-valued, then we will say that f is Riemann integrable on [a, b] ifboth g and h are Riemann integrable on [a, b] as real-valued functions. We thendefine the integral of f on [a, b] by

(2.3.2)

∫ b

af(t) dt =

∫ b

ag(t) dt + i

∫ b

ah(t) dt.



1

i

0

1+i

Figure 2.3.2. The Path of Example 2.3.4.

This Riemann integral for complex-valued functions has the properties onewould expect given knowledge of the Riemann integral for real-valued functions.The next three theorems cover some of these properties.

Theorem 2.3.5. Let f1 and f2 be Riemann integrable functions on [a, b] and αand β complex numbers. Then, αf1 + βf2 is integrable on [a, b], and

∫ b

a(αf1(t) + βf2(t)) dt = α

∫ b

af1(t) dt + β

∫ b

af2(t) dt.

Proof. That this is true if the constants α and β are real follows directly fromexpressing f1 and f2 in terms of their real and imaginary parts. Thus, to prove

the theorem we just need to show that∫ b

a if(t) dt = i∫ b

a f(t) dt if f = g + ih is anintegrable function on [a, b]. However,

∫ b

ai(g(t) + ih(t)) dt =

∫ b

a(−h(t) + ig(t)) dt

= −∫ b

ah(t) dt + i

∫ b

ag(t) dt = i

(∫ b

a(g(t) + ih(t)) dt

).

This completes the proof. !Theorem 2.3.6. If f is a function defined on [a, b] and c ∈ (a, b), then f isintegrable on [a, b] if and only if it is integrable on [a, c] and [c, b]. In this case

∫ b

af(t) dt =

∫ c

af(t) dt +

∫ b

cf(t) dt.

Proof. This follows from the fact that the same things are true of the integrals ofthe real and imaginary parts g and h of f . !Theorem 2.3.7. If f is an integrable function on [a, b], then

∣∣∣∣∣

∫ b

af(t) dt

∣∣∣∣∣ ≤∫ b

a|f(t)| dt.



Proof. This is proved using a trick. We set w =∫ b

a f(t) dt. If w = 0, there isnothing to prove. If w = 0, let u = w/|w|. Then uw = |w| and so

∣∣∣∣∣

∫ b

af(t) dt

∣∣∣∣∣ = u

∫ b

af(t) dt =

∫ b

auf(t) dt.

Since this is a real number, the integral of the imaginary part of uf is zero and wehave ∣∣∣∣∣

∫ b

af(t) dt

∣∣∣∣∣ =

∫ b

aRe(uf(t)) dt ≤

∫ b

a|uf(t)| dt =

∫ b

a|f(t)| dt. !

A complex-valued function which is defined and continuous on an interval [a, b]is clearly Riemann integrable on [a, b], since its real and imaginary parts are continu-ous, and continuous real-valued functions on closed, bounded intervals are Riemannintegrable.

Integration Along a Path. If γ is a path, then γ′ exists and is continuous oneach interval [ai−1, ai] in a partition a = a0 < a1 < · · · < an = b of the parameterinterval [a, b]. At the points a1, a2, · · · , an−1 the definition of γ′ is ambiguous –γ′(aj) has one value from the derivative of γ on [aj−1, aj ] and another from thederivative of γ on [aj , aj+1]. In order to remove this ambiguity, we choose to defineγ′ so as to be left continuous at these points. That is, at aj , we choose the value forγ′ that comes from its definition on [aj−1, aj ]. Then γ′ is well defined on I = [a, b].

If f is a complex-valued function defined and continuous on a set E containingγ(I), then the function f(γ(t))γ′(t) is a well-defined function on I which is piecewisecontinuous in the following sense: It is continuous everywhere on [a, b] except atthe partition points a1, a2, · · · , an−1. It is left continuous at these points, and thelimit from the right exists and is finite at these points as well. In other words, thisfunction is continuous from the left everywhere on [a, b] and continuous except atfinitely many points where it has simple jump discontinuities.

A function of this type is Riemann integrable on [a, b]. To see this, first observethat it is Riemann integrable on each subinterval [aj−1, aj ] because, on such aninterval, the function agrees with a continuous function except at one point, aj−1.A continuous function on a closed interval is Riemann integrable and changing itsvalue at one point does not effect this fact or the value of the integral. Furthermore,by Theorem 2.3.6, if a function is Riemann integrable on two contiguous intervals,then it is integrable on their union. It follows that a function which is integrableon each subinterval in a partition of [a, b] will be integrable on [a, b].

The above discussion settles the question of the Riemann integrability of theintegrand in the following definition.

Definition 2.3.8. Let γ : [a, b] → C be a path and let f be a function which isdefined and continuous on a set E which contains γ([a, b]). Then we define theintegral of f over γ to be

(2.3.3)

∫

γf(z) dz =

∫ b

af(γ(t))γ′(t) dt.



One may think of this definition in the following way: the contour integral onthe left in (2.3.3) is defined to be the Riemann integral obtained by replacing z byγ(t) and dz by γ′(t)dt and integrating over the parameter interval for γ.

In practice, we will calculate contour integrals by breaking the path up into itssmooth sections, calculating the integrals over these sections and then adding theresults. That this is legitimate follows from the fact that the Riemann integral ofa function over the union of two contiguous intervals on the line is the sum of theintegrals over the two intervals.

Examples.

Example 2.3.9. Find∫γ z dz if γ is the circular path defined in Example 2.3.2.

Solution: By Example 2.3.2, we have γ(t) = r eit for 0 ≤ t ≤ 2π and γ′(t) =ir eit. Thus,

∫

γz dz =

∫ 2π

0r eit ir eit dt = ir2

∫e2it dt = ir2

∫ 2π

0(cos 2t + i sin 2t) dt

= ir2

∫ 2π

0cos 2t dt − r2

∫ 2π

0sin 2t dt = 0.

Example 2.3.10. Find a path γ which traces the straight line from 0 to i followedby the straight line from i to i + 1. Then calculate

∫γ z2 dz for this path γ.

Solution: We may choose γ to be the path parameterized on [0, 2] as follows:

γ(t) =

{it, if 0 ≤ t ≤ 1;

i + t − 1, if 1 ≤ t ≤ 2.

We calculate the integrals over each of the two smooth sections of the path. On[0, 1] we have (γ(t))2 = −t2 and γ′(t) = i. Thus, the integral over the first sectionof the path is

∫ 1

0(γ(t))2γ′(t) dt =

∫ 1

0−t2i dt = −t3i/3

∣∣10

= −i/3.

On [1, 2] we have (γ(t))2 = t2 − 2t+2(t− 1)i and γ′(t) = 1. Thus, the integral overthe second section of the path is∫ 2

1(γ(t))2γ′(t) dt =

∫ 1

0(t2−2t+2(t−1)i) dt = (t3/3− t2 +(t2−2t)i)

∣∣21

= −2/3+ i.

Thus,∫γ z2 dz = −i/3 − 2/3 + i = −2/3 + 2i/3.

Example 2.3.11. Find a path γ which traces once around the triangle with vertices0, 1, i in the counterclockwise direction, starting at 0. For this path γ, find

∫γ z dz.

Solution: A path γ with the required properties has parameter interval [0, 3]and is given by

γ(t) =

⎧⎪⎨

⎪⎩

t, if 0 ≤ t ≤ 1;

2 − t + (t − 1)i if 1 ≤ t ≤ 2;

(3 − t)i if 2 ≤ t ≤ 3.



On the interval [0, 1], we have γ(t) = t and γ′(t) = 1. Hence,∫ 1

0γ(t)γ′(t) dt =

∫ 1

0t dt = 1/2.

On the interval [1, 2], we have γ(t) = 2 − t − (t − 1)i and γ′(t) = −1 + i. Hence,∫ 2

1γ(t)γ′(t) dt =

∫ 2

1(2t − 3 + i) dt = i.

On the interval [2, 3], we have γ(t) = t − 3 and γ′(t) = −i. Hence,∫ 1

0γ(t)γ′(t) dt =

∫ 3

2(3 − t)i dt = i/2.

If we add the contributions of each of these three intervals, the result is∫

γz dz = 1/2 + i + i/2 = 1/2 + (3/2)i.

Exercise Set 2.3

1. Find∫ π0 eit dt.

2. Find∫ 10 sin(it) dt.

3. Find∫ 2π0 eint eimt dt for all integers n and m.

4. Find a path which traces the straight line joining 2 − i to −1 + 3i.

5. If z0 ∈ C, find a path which traces the circle of radius r, centered at z0, (a)once in the counterclockwise direction, (b) once in the clockwise direction, (c)three times in the counterclockwise direction.

6. Prove that if γ(t) = x(t) + iy(t) is a curve defined on an interval I, with realand imaginary parts x(t) and y(t), and if c ∈ I, then γ′(c) exists if and only ifx′(c) and y′(c) exist and, in this case, γ′(c) = x′(c) + iy′(c).

7. Show that if f is a smooth complex-valued function on an interval [a, b], then∫ ba f ′(t) dt = f(b) − f(a).

8. Suppose γ is a path with parameter interval [a, b]. Use the result of the previousexercise to show that

∫γ 1 dz = γ(b) − γ(a).

9. Find∫γ z2 dz if γ traces a straight line from 0 to w.

10. Find∫γ z−1 dz and

∫γ z dz for the circular path γ(t) = 3 eit, 0 ≤ t ≤ 2π.

11. Find∫γ Re(z) dz if γ is the path of Example 2.3.11.

12. With γ as in the previous exercise, find∫γ Im(z2) dz.

13. Is it generally true that Re(∫γ f(z) dz) =

∫γ Re(f(z)) dz?


2.4. Properties of Contour Integrals 47

2.4. Properties of Contour Integrals

We begin this section with the question of parameter independence. To whatextent does the integral of a function along a path depend on how the path isparameterized? The same geometric figure γ(I) may be parameterized in manyways. For example, the top third of the unit circle may be parameterized by

γ1(t) = −t + i√

1 − t2, −√

3/2 ≤ t ≤√

3/2, or

γ2(t) = eit = cos t + i sin t, π/6 ≤ t ≤ 5π/6,(2.4.1)

and these are only two of infinitely many possibilities. Does the integral of a functionover the upper third of the unit circle depend on which of these parmeterizationsis chosen?

Parameter Changes that Change the Integral. The following example showsthat some changes of parameterization do change the integral.

Example 2.4.1. Find∫γ1

1/z dz if γ1(t) = r eit on [0, 2π] is the circular path ofExample 2.3.2. Does the answer change if the circle is traversed in the clockwisedirection instead, using the path γ2(t) = r e−it on [0, 2π]?

Solution: From Example 2.3.2 we know that the path γ1(t) = r eit has γ′1(t) =

ir eit and so the given integral is∫

γ1

dz

z=

∫ 2π

0

(r eit)′

r eitdt =

∫ 2π

0i dt = 2πi.

On the other hand, the derivative of γ2 = e−it is −ir e−it and so∫

γ2

dz

z=

∫ 2π

0−i dt = −2πi.

This example shows that the integral along a path depends not only on thegeometric figure that is the image γ(I) of the path, but also on the direction thepath is traversed (at the very least).

Also, traversing a portion of the curve more than once may affect the integral.For example, if we were to go around the circle twice in Example 2.4.1, by choosingγ(t) = e2it on [0, 2π], the result would be 4πi instead of 2πi.

The Independence of Parameterization Theorem. There is a degree to whichthe integral is independent of the parameterization. Certain ways of changing theparameterization do not effect the integral, as the following theorem shows.

Theorem 2.4.2. Let γ1 : [a, b] → C be a path and α : [c, d] → [a, b] a smoothfunction with α(c) = a and α(d) = b. If γ2 is the path with parameter interval [c, d]defined by γ2(t) = γ1(α(t)), then

∫

γ2

f(z)dz =

∫

γ1

f(z)dz

for every function f defined and continuous on a set E containing γ1([a, b]) =γ2([c, d]).



Proof. We have γ2(t) = γ1(α(t)) and, by the chain rule,

γ′2(t) = γ′

1(α(t))α′(t).

Thus,∫

γ2

f(z) dz =

∫ d

cf(γ2(t))γ

′2(t) dt

=

∫ d

cf(γ1(α(t)))γ′

1(α(t))α′(t) dt

=

∫ b

af(γ1(s))γ

′1(s) ds

=

∫

γ1

f(z) dz,

where the third equality follows from the substitution s = α(t). This completes theproof. !

Note that the condition that α(c) = a and α(d) = b is essential in the abovetheorem. It says that α takes the endpoints of the parameter inverval [c, d] to theendpoints of the parameter interval [a, b] in an order preserving fashion.

Example 2.4.3. Are the integrals of a continuous function over the two paths in(2.4.1) necessarily the same?

Solution: Yes. If we set α(t) = − cos t, then α is a smooth function mappingthe parameter interval [π/6, 5π/6] to the parameter interval [−

√3/2,

√3/2] in an

order preserving fashion. Furthermore, γ2 = γ1 ◦ α. Thus, the above theoreminsures that the integral of a continuous function over γ1 is the same as its integralover γ2.

Doesn’t Example 2.4.1 contradict Theorem 2.4.2? After all, if γ1(t) = eit on[0, 2π] and α : [0, 2π] → [0, 2π] is defined by α(t) = 2π − t, then γ2(t) = γ1(α(t)) =e−it. By Example 2.4.1 the integrals of 1/z over these two curves are different.Doesn’t Theorem 2.4.2 say they should be the same? No. The conditions α(a) = cand α(b) = d are not satisfied by this choice of α, since α(0) = 2π and α(2π) = 0.In other words, this choice of α reverses the order of the endpoints of the parameterinterval rather than preserving that order.

In general, the conditions α(a) = c and α(b) = d guarantee that, overall, γ2

traverses the curve in the same direction as γ1. If α′ were positive on the entireinterval, then α would be increasing on this interval and γ1 and γ2 would be movingin the same direction at each point of the curve. If α′ is not positive on all of [c, d],then there may be intervals where one path reverses direction and backtracks, whilethe other path does not. These things do not affect the integral, because if a curvedoes backtrack for a time, it has to turn around and recover the same ground inorder to catch up to the other curve in the end. This is an intuitive explanation;the actual proof that the integral is unaffected is in the proof of the above theorem.

Theorem 2.4.2 leads to a strategy which, for some paths γ1 and γ2 with thesame image, yields a proof that they determine the same integral: Suppose that theparameter intervals for the two paths can each be partitioned into n subintervals



in such a way that for j = 1, · · · , n, γ1 on its jth subinterval and γ2 on its jthsubinterval are related by a smooth function αj , as in Theorem 2.4.2. If this canbe done, then it clearly follows that

∫γ1

f(z) dz =∫γ2

f(z) dz for any function f

which is continuous on a set containing γ1(I). For this reason, Theorem 2.4.2 issometimes called the independence of parameterization theorem.

Remark 2.4.4. Since path integrals are essentially independent of the way thepath is parameterized, we will often describe a path without specifying a parame-terization. Instead, we will just give a description of the geometric object that istraced, the direction, and how many times. For example, we may describe a pathas tracing once around the unit circle in the counterclockwise direction, or tracingonce around the boundary ∂∆ of a given triangle ∆ in the counterclockwise direc-tion, or as tracing the straight line path from a complex number w1 to a complexnumber w2. In the first two cases we may simply write

∫

|z|=1f(z) dz or

∫

∂∆f(z) dz

for the corresponding path integral. In the latter case, we may write∫ w2

w1

f(z) dz

for the path integral along the straight line from w1 to w2.

Closed Curves. The curves in Examples 2.3.2 and 2.3.4 both have the propertythat they begin and end at the same point – that is, they are closed curves. A closedcurve γ on a parameter interval [a, b] is one that satisfies γ(a) = γ(b). A closedcurve which is a path will be called a closed path.

The famous integral theorem of Cauchy states that the integral of an analyticfunction f around a closed path is 0, provided there is an appropriate relationshipbetween the curve γ and the domain U on which f is analytic (roughly speaking,the curve should lie in U but not go around any holes in U). Since the functionf(z) = z is analytic on C (as is any polynomial in z), the next example illustratesthis phenomenon.

Example 2.4.5. Find∫γ z dz if γ is the path of Example 2.3.4.

Solution: From Example 2.3.4 we know that the path γ(t) has values 4t, 1 +(4t−1)i, 3−4t+i, (4−4t)i and derivatives 4, 4i,−4, and −4i on the four subintervalsof the partition 0 < 1/4 < 1/2 < 3/4 < 1. Thus, the integrals over the four smoothpieces of our curve are

∫ 1/4

04t · 4 dt = 8t2

∣∣1/4

0= 1/2,

∫ 1/2

1/4(1 + (4t − 1)i) · 4i dt = (4ti − 8t2 + 4t)

∣∣1/2

1/4= i − 1/2,

∫ 3/4

1/2(3 − 4t + i) · (−4) dt = (−12t + 8t2 − 4ti)

∣∣3/4

1/2= −1/2 − i,

∫ 1

3/4(4 − 4t)i · (−4i) dt = (+16t − 8t2)

∣∣13/4

= 1/2.



a cb

γ (a)1 2γ1γ2γ (c)

γ (b)1 2γ (b)=

Figure 2.4.1. The Join of Two Paths.

Since these add up to 0, we have∫γ z dz = 0.

The function 1/z is also analytic, except at z = 0. The circular path of Example2.4.1 is closed and lies in the domain where 1/z is analytic. So why is the integralnot 0? Because the path goes around a hole in the domain of 1/z – it goes around{0}.

Additivity Properties of Contour Integrals. If γ is a path with parameterinterval [a, b], then we can use Theorem 2.4.2 to change the parameter interval toany other interval [c, d] with c < d, in a way that does not affect the image of γ orintegrals over γ. In fact, if we set

α(t) = a +b − a

d − c(t − c),

then α is smooth, α([c, d]) = [a, b], α(c) = a and α(d) = b. Thus, γ1(t) = γ(α(t))defines a path γ1 with the same image as γ and, by Theorem 2.4.2, a path whichdetermines the same integral for continuous functions on its image. Thus, withoutloss of generality, we may always assume that the parameter interval for a path isany interval we choose.

If γ1 and γ2 are two paths so that γ1 ends where γ2 begins, then we can jointhe two paths to form a single new path γ1 + γ2. We do this as follows: If γ1 hasparameter interval [a, b], we choose a parameter interval of the form [b, c] for γ2.The fact that γ2 begins where γ1 ends means that γ1(b) = γ2(b). We define γ1 + γ2

on [a, c] by

(2.4.2) (γ1 + γ2)(t) =

{γ1(t) if t ∈ [a, b],

γ2(t) if t ∈ [b, c].

The path γ1 + γ2 is called the join of γ1 and γ2.

In Example 2.4.1, changing the path from one tracing the circle counterclock-wise to one tracing the circle clockwise had the effect of changing the sign of theintegral. As we shall see, this always happens. If γ : [a, b] :→ C is a path, denoteby −γ the path defined by

−γ(t) = γ(a + b − t).



Then −γ(a) = γ(b) and −γ(b) = γ(a). In fact, −γ traces the same geometric figureas γ, but it does so in the opposite direction.

For some closed curves, such as circles, and boundaries of rectangles, triangles,etc., there is clearly a clockwise direction around the curve and a counterclockwisedirection. If such a curve is parameterized so that it is traversed in the counter-clockwise direction, we will say the resulting closed path has positive orientation.If it is traversed in the clockwise direction, we will say it has negative orientation.Clearly, if γ has positive orientation, then −γ has negative orientation.

The common starting and ending point of a closed path can be changed withoutchanging the integral of a function over this path. This is done by representing theclosed path as the join of two paths which connect the original starting and endingpoint to the new one. One then uses part (b) of the next theorem. The details areleft to the exercises.

The next theorem states the elementary properties of path integrals havingto do with linearity and path additivity. Part (b) follows immediately from thecorresponding additivity property of the Riemann integral on the line and we havealready used it several times. We leave the proofs of (a) and (c) to the exercises(Exercise 2.4.5).

Theorem 2.4.6. Let γ, γ1, γ2 be paths with γ1 ending where γ2 begins, f and g twofunctions which are continuous on a set E containing the images of these paths,and a and b complex numbers. Then

(a)

∫

γ(af(z) + bg(z)) dz = a

∫

γf(z) dz + b

∫

γg(z) dz;

(b)

∫

γ1+γ2

f(z) dz =

∫

γ1

f(z) dz +

∫

γ2

f(z) dz;

(c)

∫

−γf(z) dz = −

∫

γf(z) dz.

Part (a) of this theorem says that a path integral is a linear function of theintegrand, Part (b) says that it is an additive function of the path, while Part (c)shows why the notation −γ is appropriate for the curve that is γ traversed in theopposite direction.

Length of a Path. We define the length ℓ(γ) of a path γ in C in the same waythe length of a curve in R2 is defined in calculus.

Definition 2.4.7. If γ(t) = x(t) + iy(t) is a path in C with parameter interval[a, b], then the length ℓ(γ) of γ is defined to be

ℓ(γ) =

∫ b

a|γ′(t)| dt =

∫ b

a

√x′(t)2 + y′(t)2 dt.

Example 2.4.8. Prove that the above definition of length yields the correct lengthfor a path which traces once around a circle of radius r.

Solution: The path is γ(t) = r eit, with parameter interval [0, 2π]. The deriv-

ative of γ is γ′(t) = ir eit and so |γ′(t)| = r. Thus, ℓ(γ) =∫ 2π0 r dt = 2πr.



It will be important in coming sections to be able to obtain good upper boundson the absolute value of a path integral. The key theorem that produces such upperbounds is the following.

Theorem 2.4.9. Let γ be a path in C and f a function continuous on a set con-taining γ(I). If |f(z)| ≤ M for all z ∈ γ(I), then

∣∣∣∣

∫

γf(z) dz

∣∣∣∣ ≤ Mℓ(γ).

Proof. If the parameter interval for γ is [a, b], then∣∣∣∣∫

γf(z) dz

∣∣∣∣ =

∣∣∣∣∣

∫ b

af(γ(t))γ′(t) dt

∣∣∣∣∣ ≤∫ b

a|f(γ(t))γ′(t)| dt

≤∫ b

aM |γ′(t)| dt = M

∫ b

a|γ′(t)| dt = Mℓ(γ). !

The next example is a typical application of this theorem.

Example 2.4.10. Show that if f is a bounded continuous function on C, and γR

is the path γR(z) = R eit for t ∈ [0, 2π], then

(2.4.3) limR→∞

∫

γR

f(z)

(z − w)2dz = 0

for each w ∈ C.

Solution: The statement that f is bounded means there is an upper boundM for |f |. That is, |f(z)| ≤ M for all z ∈ C. We also have |z − w| ≥ |z| − |w|by the second form of the triangle inequality. If z ∈ γ(I), then |z| = R and so|z − w| ≥ R − |w|, which implies |z − w|−2 ≤ (R − |w|)−2. Thus, for z ∈ γ(I), wehave the following bound on the integrand of (2.4.3):

∣∣∣∣f(z)

z − w

∣∣∣∣ ≤M

(R − |w|)2 .

Since ℓ(γR) = 2πR, Theorem 2.4.9 implies that∣∣∣∣∫

γR

f(z)

(z − w)2dz

∣∣∣∣ ≤2πMR

(R − |w|)2 .

The right side of this inequality has limit 0 as R → ∞ and this implies (2.4.3).

Exercise Set 2.4

1. Compute∫γ z2 dz if γ is any path which traces once around the circle of radius

one in the counterclockwise direction.

2. Compute∫γ 1/z dz if γ is any path which traces twice around the circle of radius

one, centered at 0, in the counterclockwise direction.

3. If z0 and w0 are two points of C, compute∫γ z dz if γ is any path which traces

the straight line from z0 to w0 once.


2.5. Cauchy’s Integral Theorem for a Triangle 53

4. Compute the integral of the previous exercise for any smooth path γ whichbegins at z0 and ends at w0.

5. Prove Parts (a) and (c) of Theorem 2.4.6.

6. Describe a smooth, order preserving function α which takes the parameterinterval [0, 1] to the parameter interval [2, 5].

7. Prove that a parameter change γ → γ ◦ α, like the one in Theorem 2.4.2, doesnot change the length of a path provided α is an non-decreasing function (hasa non-negative derivative).

8. Show that

∣∣∣∣∫

γ

cos z

zdz

∣∣∣∣ ≤ 2πe if γ is a path that traces the unit circle once.

Hint: Show that | cos z| ≤ e if |z| = 1.

9. Show that if ∆ is a triangle in the plane of diameter d (length of its longestside), and if f is a continuous function on ∆ with |f | bounded by M on ∆,then ∣∣∣∣

∫

∂∆f(z) dz

∣∣∣∣ ≤ 3Md.

10. Prove that∫γ p(z) dz = 0 if γ(t) = eit, 0 ≤ t ≤ 2π, and p(z) is any polynomial

in z (this is a special case of Cauchy’s Theorem, but do not assume Cauchy’sTheorem in your proof).

11. Let R(z) be the remainder after n terms in the power series for ez. That is,

R(z) = ez −n∑

k=1

zk

k!=

∞∑

k=n+1

zk

k!.

Prove that |R(z)| ≤ e − 1

(n + 1)!if |z| ≤ 1.

12. Prove that∫γ ez dz = 0 if γ(t) = eit, 0 ≤ t ≤ 2π using the previous exercise

and Exercise 10.

13. Prove that if γ is a closed path with parameter interval I = [a, b] and commonstarting and ending point z = γ(a) = γ(b) and w is any other point on γ(I),then there is another closed path γ1 with γ1(I) = γ(I), which determines thesame integral, but has w as common starting and ending point. Hint: Use part(b) of Theorem 2.4.6.

2.5. Cauchy’s Integral Theorem for a Triangle

The core material of any beginning Complex Variables text is the proof of Cauchy’sIntegral Theorem and the exploration of its consequences. Roughly speaking,Cauchy’s Integral Theorem states that the integral of an analytic function arounda closed path is zero, provided the path is contained in the open set U on which thefunction is analytic and does not go around any “holes” in U . Part of the problemhere is to make sense of the idea of a “hole” in an open set and to decide what itmeans for a path to go around such a hole.


2.5. Cauchy’s Integral Theorem for a Triangle 59

c

c

(a) (b)

Figure 2.5.3. Dealing with an Exceptional Point c.

of each of these is zero, so the integral around ∂∆ is also 0. This completes theproof. !

Exercise Set 2.5

1. Prove the Bolzano-Weierstrass Theorem: If K is a compact subset of Rn, thenevery sequence in K has a subsequence which converges to an element of K.

2. Use Corollary 2.5.5 to show that if K is a compact subset of C and f is acontinuous complex-valued function on K, then the modulus |f(z)| of f takeson a maximal value at some point of K.

3. Show that if K is a compact subset of C, then there is a point z0 ∈ K ofminimum modulus – that is, a point z0 ∈ K such that

|z0| ≤ |z| for all z ∈ K.

4. Prove that if g is analytic on an open subset U of C and γ : [a, b] → U is a pathin U , then

(g(γ(t)))′ = g′(γ(t))γ′(t)

for t ∈ [a, b]. Hint: The proof is very similar to the proof of Theorem 2.2.7.

5. Calculate∫γ zn dz if n is a non-negative integer and γ is a path in the plane

joining the point z0 to the point w0. Hint: Use Theorem 2.5.6.

6. Show that∫γ p(z) dz = 0 if γ is any closed path in the plane and p is any

polynomial.

7. Calculate∫γ 1/z dz if γ is any path in C joining −i to i which does not cross

the half-line (−∞, 0] on the real axis. Hint: Use the result of Example 2.2.11and Theorem 2.5.6.



8. Using the same hint as in the previous exercise, show that∫

γ

1

zdz = 0

if γ is any closed path contained in the complement of the set of non-positivereal numbers. Compare this with Example 2.4.1.

9. If√

z is defined by√

z = e(log z)/2 for the branch of the log function definedby the condition −π/2 ≤ arg(z) ≤ 3π/2, find an antiderivative for

√z and

then find∫γ

√z dz, where γ is any path from −1 to 1 which lies in the upper

half-plane.

10. Prove that if f is analytic in an open set containing a rectangle R, then thepath integral of f around the boundary of this rectangle is 0.

11. Let γ be the path which traces the straight line from 1 to 1+i, then the straightline from 1 + i to i and then the straight line from i to 0. Calculate

∫γ zn dz.

12. Let ∆ be the triangle with vertices 1− i, i, and−1− i and S be the square withvertices 1− i, 1 + i,−1 + i, and− 1− i. If f is any function which is analytic onC \ {0}, prove that ∫

∂∆f(z) dz =

∫

∂Sf(z) dz,

where ∂∆ and ∂S are traversed in the counterclockwise direction.

13. For any pair of points a, b in C, denote the integral of a function f along the

straight line segment joining a to b by∫ b

a f(z) dz, as in Remark 2.4.4. Supposef is analytic in an open set containing the triangle with vertices a, b, c. Showthat ∫ c

af(z) dz −

∫ b

af(z) dz =

∫ c

bf(z) dz.

14. Show that Theorem 2.5.9 can be strengthened to conclude that the integral off around any triangle in U is 0 if f is continuous on U and analytic on U \ I,where I is an interval contained in U . Hint: First consider the case where oneside of the triangle lies along the interval I.

15. If f is analytic on an open set U , then the integral of f around the boundary ofany triangle in U is 0 (Theorem 2.5.8), as is its integral around the boundaryof any rectangle in U (Exercise 2.5.10). What other geometric figures have thisproperty? What is the most general theorem along these lines you can thinkof?

2.6. Cauchy’s Theorem for a Convex Set

A convex set C in C is a set with the property that if a and b are points in C, thenthe line segment joining a and b is also contained in C.

Existence of Antiderivatives. The strategy for proving Cauchy’s Theorem forconvex sets is to prove that every analytic function on a convex set has an antideriv-ative and then apply Theorem 2.5.6. The first step in this program is accomplishedwith the following theorem.


2.6. Cauchy’s Theorem for a Convex Set 65

even if the function is not analytic at some point but is continuous there, it followsthat

0 =

∫

γg(z, w)dw =

∫

γ

f(w)

w − zdw −

∫

γ

f(z)

w − zdw

=

∫

γ

f(w)

w − zdw − 2πi Indγ(z)f(z),

as long as z is not on the contour γ (note that this is required in order to writethe integral in the first line above as the difference of two integrals, since otherwisethese two integrals might not exist individually). We conclude that

Indγ(z)f(z) =1

2πi

∫

γ

f(w)

w − zdw,

as required. This completes the proof. !

This is a striking result, for it says that the values of an analytic function atpoints “inside” a closed path are determined by its values at points on the path.Here, a point is considered inside the path if the path has non-zero index at thepoint.

Corollary 2.6.8. If U is a convex open set, z ∈ U and γ is a closed path in Uwith Indγ(z) = 1, then

f(z) =1

2πi

∫

γ

f(w)

w − zdw

for every function f analytic on U .

Intuitively, the meaning of the hypothesis Indγ(z) = 1 in the above corollary isthat the closed path γ goes around z once and does so in the positive direction.

Cauchy’s Integral Theorem and Cauchy’s Integral Formula have a wealth ofapplications. We will begin exploring these in the next chapter.

However, in order for Cauchy’s Integral Theorem, in the above form, to be us-able, we need to be able to easily compute the index of a curve around a given point.The last section of this chapter is devoted to developing the essential properties ofthe index function which make this possible.

Exercise Set 2.6

1. Prove that a function which has complex derivative identically 0 on a convexopen set U is constant on U .

2. Calculate∫γ(z2 − 4)−1 dz if γ is the unit circle traversed once in the positive

direction.

3. Calculate∫γ(1 − ez)−1 dz if γ is the circle γ(t) = 2i + eit.

4. Calculate∫γ 1/z dz if γ is any circle which does not pass through 0. Note that

the answer depends on γ.

5. Find∫γ 1/z2 dz if γ is any closed path in C \ {0}.



6. Show that the principal branch of the log function can be described by theformula log(z) =

∫ z1 1/w dw for z /∈ (−∞, 0].

7. Prove Corollary 2.6.3.

8. Without doing any calculating, show that the integral of 1/z around the bound-ary of the triangle with vertices i, 1 − i,−1 − i is 2πi.

9. Let f be a function which is analytic on C\{z0}. Show that the contour integralof f around a circle of radius r > 0, centered at z0, is independent of r.

10. Calculate Indγ(z0) if γ(t) = z0 + eint, t ∈ [0, 2π] and n is any integer.

11. Calculate Indγ(1 + i) if γ is the path which traces the line from 0 to 2, thenproceeds counterclockwis around the circle |z| = 2 from 2 to 2i and then tracesthe line from 2i to 0. What is the answer if this path is traversed in the oppositedirection?

12. Use Cauchy’s Integral Formula to calculate

∫

|z|=1

ez

zdz.

13. Use Cauchy’s Formula to show that∫

|z−1|=1

1

z2 − 1dz = πi,

∫

|z+1|=1

1

z2 − 1dz = −πi.

14. Show that ∫

|z|=3

1

z2 − 1dz = 0.

Hint: Use the result of the preceding exercise.

15. Use Cauchy’s Integral Formula to prove that if f is a function which is analyticin an open set containing the closed unit disc D1(0), and if T = {z : |z| = 1} isthe unit circle, then |f(0)| ≤ M , where M is the maximum value of |f | on T .

16. Show that if γ is a path from z1 to z2 which does not pass through the pointz0, then

∫

γ

1

w − z0dw = log

(z2 − z0

z1 − z0

),

for some branch of the log function. Note that, in the case where z1 = z2, thisis just Theorem 2.6.6.

2.7. Properties of the Index Function

If γ is a closed path, then removing γ(I) from the plane results in a set which isdivided into a number of connected pieces. These are open sets called the connectedcomponents of the complement of γ(I). We will prove that Indγ(z) is constant oneach of these components. Thus, to calculate Indγ(z) on a given component, oneonly needs to calculate it at one point of the component.

Before proving this, we need to have a firm idea of what a connected componentis. This leads to a discussion of connected sets.


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